The equilibrium constant for the reaction is is 9.1 x 10-6(aq) at 298 K

Question

The equilibrium constant for the reaction is is 9.1 x 10^-6(aq) at 298 K when

2 Fe³⁺(aq) + Hg₂²⁺(aq) <=> 2 Fe²⁺(aq) + 2 Hg²⁺(aq)

Calculate ΔG in J when

{Fe³⁺(aq)} = 0.725

{Hg₂²⁺(aq)} = 0.022

{Fe²⁺(aq)} = 0.033

{Hg²⁺(aq)} = 0.056

I found Kc = (.033)²(.056)² / (.725)²(.022)

= 2.95 x 10^-4

ΔG = (-8.3145)(298)(ln(2.95x10^-4)

= 20140.24

but the answer is 8621.5

please list the steps in the proper solution so I can follow along and see where I went wrong

Thank you!

Answer 1

Well, your calculation of K is correct, but it isn't really K, because K is given as 9.1x10^-6. What you have correctly calculated is Q. So, what you need to do is find ∆Gº using the given value of K, and then use

∆G = ∆Gº + RT ln Q

So, ∆Gº = - RT ln K = - (8.314)(298)(ln 9.1x10^-6) = 28,739 J

Then use this value and your value of Q to solve for ∆G...

∆G = ∆Gº + RT ln Q

∆G = 28,739 + (8.314)(298)(ln 2.95x10-4) = 28,739 - 20,143 = 8596 J (close to the correct answer but off maybe due to rounding or a math error. But this is the general idea of how to do the problem). Hope this helps.